JEE Mathematics 2024 Question with Solution

Given: $\frac{dx}{dt}+ax=0$, $\frac{dy}{dt}+by=0$, $x(0)=2$, $y(0)=1$, and $3y(1)=2x(1)$.

Find: The value of $t$ for which $x(t)=y(t)$.

Solving the differential equations gives

$$x(t)=x(0)e^{-at}=2e^{-at}$$

and

$$y(t)=y(0)e^{-bt}=e^{-bt}$$

Using $3y(1)=2x(1)$,

$$3e^{-b}=2\cdot 2e^{-a}=4e^{-a}$$

Taking logarithm,

$$-b+\log 3=-a+\log 4$$

So,

$$a-b=\log\left(\frac{4}{3}\right)$$

Now set $x(t)=y(t)$:

$$2e^{-at}=e^{-bt}$$

Taking logarithm,

$$\log 2-at=-bt$$

Hence,

$$t(a-b)=\log 2$$

Substituting $a-b=\log\left(\frac{4}{3}\right)$,

$$t\log\left(\frac{4}{3}\right)=\log 2$$

Therefore,

$$t=\frac{\log 2}{\log\left(\frac{4}{3}\right)}$$

the solution concludes this as $t=\log_4\left(\frac{2}{3}\right)$, so the correct option on the solution's is D. There is a simplification discrepancy in the source working, but the extracted answer from the solution is D.

Given: $x(t)=2e^{-at}$ and $y(t)=e^{-bt}$ after solving the two first-order differential equations.

Find: The time when $x(t)=y(t)$.

From the condition at $t=1$,

$$3y(1)=2x(1)$$

so

$$3e^{-b}=2\left(2e^{-a}\right)=4e^{-a}$$

Rearranging,

$$e^{a-b}=\frac{4}{3}$$

Taking logarithm,

$$a-b=\log\left(\frac{4}{3}\right)$$

Now, for $x(t)=y(t)$,

$$2e^{-at}=e^{-bt}$$

which gives

$$2=e^{(a-b)t}$$

or equivalently,

$$\log 2=t(a-b)$$

Substitute the value of $a-b$:

$$\log 2=t\log\left(\frac{4}{3}\right)$$

Thus,

$$t=\frac{\log 2}{\log\left(\frac{4}{3}\right)}$$

The source solution marks option D as correct and states the final answer as $\log_4\left(\frac{2}{3}\right)$. Therefore, the correct option according to the solution is D.